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Fourier's law: $dQ/dt = kA\Delta T/L$ for steady-state conduction through a uniform slab. Thermal conductivity $k$ (W m$^{-1}$ K$^{-1}$) ranges from silver (429) to styrofoam (0.01). The thermal resistance analogy maps directly to electrical circuits: $R = L/(kA)$, $dQ/dt = \Delta T/R$.

For composite walls in series: $R_{\text{total}} = R_1 + R_2 + ...$, same heat flow through all layers. The interface temperature: $T_i = T_{\text{hot}} - (dQ/dt) \times R_1$. For parallel paths: $1/R_{\text{total}} = 1/R_1 + 1/R_2$, same temperature difference across all paths.

Effective conductivity formulas: Series (equal thicknesses): $k_{\text{eff}} = 2k_1k_2/(k_1+k_2)$ (harmonic mean). Parallel (equal areas): $k_{\text{eff}} = (k_1+k_2)/2$ (arithmetic mean). In steady state with no lateral losses, temperature varies linearly along a uniform rod.

A thin insulation layer can be remarkably effective — its low $k$ creates high resistance despite small thickness. This is why fiberglass insulation (a few centimeters) dramatically reduces building heat loss.